Snakemake使用所有样本作为porechop的一个输入

Question

我正在尝试通过Snakemake工作流对几个数据使用porechop。

在我的Snakefile中，除了all规则之外，还有三个规则，一个fastqc规则和一个porechop规则。fastqc规则工作得很好，我的三个fastq都用到了。但对于porechop，它不是运行命令三次，而是同时为所有三个文件运行一次带有-i标志的命令：

Error in rule porechop:
    jobid: 2
    output: /ngs/prod/nanocea_project/test/prod/porechop/25022021_2_pore.fastq.gz, /ngs/prod/nanocea_project/test/prod/porechop/02062021_1_pore.fastq.gz, /ngs/prod/nanocea_project/test/prod/porechop/02062021_2_pore.fastq.gz
    conda-env: /ngs/prod/nanocea_project/test/.snakemake/conda/a72fb141b37718b7c37d9f32d597faeb
    shell:
        porechop -i /ngs/prod/nanocea_project/test/reads/25022021_2.fastq.gz /ngs/prod/nanocea_project/test/reads/02062021_1.fastq.gz /ngs/prod/nanocea_project/test/reads/02062021_2.fastq.gz -o /ngs/prod/nanocea_project/test/prod/porechop/25022021_2_pore.fastq.gz /ngs/prod/nanocea_project/test/prod/porechop/02062021_1_pore.fastq.gz /ngs/prod/nanocea_project/test/prod/porechop/02062021_2_pore.fastq.gz -t 40 --discard_middle
        (one of the commands exited with non-zero exit code; note that snakemake uses bash strict mode!)

但是，当我将它与单个样本一起使用时，程序可以正常工作。

下面是我的代码：

import glob
import os

###Global Variables###

FORMATS=["zip", "html"]
DIR_FASTQ="/ngs/prod/nanocea_project/test/reads"

###FASTQ Files###

def list_samples(DIR_FASTQ):
        SAMPLES=[]
        for file in glob.glob(DIR_FASTQ+"/*.fastq.gz"):
                base=os.path.basename(file)
                sample=(base.replace('.fastq.gz', ''))
                SAMPLES.append(sample)
        return(SAMPLES)

SAMPLES=list_samples(DIR_FASTQ)

###Rules###
rule all:
        input:
                expand("/ngs/prod/nanocea_project/test/stats/fastqc/{sample}_fastqc.{ext}", sample=SAMPLES, ext=FORMATS),
                expand("/ngs/prod/nanocea_project/test/prod/porechop/{sample}_pore.fastq.gz", sample=SAMPLES)
rule fastqc:
        input:
                expand(DIR_FASTQ+"/{sample}.fastq.gz", sample=SAMPLES)
        output:
                expand("/ngs/prod/nanocea_project/test/stats/fastqc/{sample}_fastqc.{ext}", sample=SAMPLES, ext=FORMATS)
        threads:
                16
        conda:
                "envs/fastqc.yaml"
        shell:
                "fastqc {input} -o /ngs/prod/nanocea_project/test/stats/fastqc/ -t {threads}"

rule porechop:
        input:
                expand(DIR_FASTQ+"/{sample}.fastq.gz", sample=SAMPLES)
        output:
                expand("/ngs/prod/nanocea_project/test/prod/porechop/{sample}_pore.fastq.gz", sample=SAMPLES)
        threads:
                40
        conda:
                "envs/porechop.yaml"
        shell:
                "porechop -i {input} -o {output} -t {threads} --discard_middle"

你知道出什么事了吗？

谢谢！

Answer 1

这个问题经常被提出来。如果您在input:或output:中使用expand()，那么您将向规则提供所有文件的列表。这和写作是一样的：

input:
    ['sample1.fastq', 'sample2.fastq', ..., 'sampleN.fastq'],
output:
    ['sample1.pore.fastq', 'sample2.pore.fastq', ..., 'sampleN.pore.fastq'],

要在每个输入/输出上运行规则，只需删除展开：

rule porechop:
    input:
        DIR_FASTQ+"/{sample}.fastq.gz"
    output:               
        "/ngs/prod/nanocea_project/test/prod/porechop/{sample}_pore.fastq.gz",